1) How to check if a String contains only numeric digits? (solution)
If you'll be processing the number as text, then change:
if (text.contains("[a-zA-Z]+") == false && text.length() > 2){
to:
if (text.matches("[0-9]+") && text.length() > 2) {
Instead of checking that the string doesn't contain alphabetic characters, check to be sure it contains only numerics.
If you actually want to use the numeric value, use Integer.parseInt() or Double.parseDouble() as others have explained below.
2) How to write LRU cache in Java using Generics? (answer)
We need to keep a pointer to the LRU and MRU items. The entries' values will be stored in the list and when we we query the HashMap, we will get a pointer to the list. On get(), we need to put the item at the right-most side of the list. On put(key,value), if the cache is full, we need to remove the item at the left-most side of the list from both the list and the HashMap.
import java.util.LinkedHashMap;
import java.util.Iterator;
public class LRUCache {
private int capacity;
private LinkedHashMap<Integer,Integer> map;
public LRUCache(int capacity) {
this.capacity = capacity;
this.map = new LinkedHashMap<>();
}
public int get(int key) {
Integer value = this.map.get(key);
if (value == null) {
value = -1;
} else {
this.set(key, value);
}
return value;
}
public void set(int key, int value) {
if (this.map.containsKey(key)) {
this.map.remove(key);
} else if (this.map.size() == this.capacity) {
Iterator<Integer> it = this.map.keySet().iterator();
it.next();
it.remove();
}
map.put(key, value);
}
}
3) Write a Java program to convert bytes to long? (answer)
public byte[] longToBytes(long x) {
ByteBuffer buffer = ByteBuffer.allocate(Long.BYTES);
buffer.putLong(x);
return buffer.array();
}
public long bytesToLong(byte[] bytes) {
ByteBuffer buffer = ByteBuffer.allocate(Long.BYTES);
buffer.put(bytes);
buffer.flip();//need flip
return buffer.getLong();
}
4) How to reverse a String in Java without using StringBuffer? (solution)
public class ReverseString {
private static String hello = "Hello World";
public static void main(String[] args)
{
System.out.println(reverseString(hello));
}
public static String reverseString(String s) {
char c[] = s.toCharArray();
int i = 0, j = c.length - 1;
while (i < j) {
char tmp = c[i];
c[i] = c[j];
c[j] = tmp;
i++;
j--;
}
return new String(c);
}
5) How to find the word with the highest frequency from a file in Java? (solution)
import java.io.BufferedReader;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.regex.Pattern;
/**
* Java program to find count of repeated words in a file.
*
* @author
*/
public class Problem {
public static void main(String args[]) {
Map<String, Integer> wordMap = buildWordMap("C:/temp/words.txt");
List<Entry<String, Integer>> list = sortByValueInDecreasingOrder(wordMap);
System.out.println("List of repeated word from file and their count");
for (Map.Entry<String, Integer> entry : list) {
if (entry.getValue() > 1) {
System.out.println(entry.getKey() + " => " + entry.getValue());
}
}
}
public static Map<String, Integer> buildWordMap(String fileName) {
// Using diamond operator for clean code
Map<String, Integer> wordMap = new HashMap<>();
// Using try-with-resource statement for automatic resource management
try (FileInputStream fis = new FileInputStream(fileName);
DataInputStream dis = new DataInputStream(fis);
BufferedReader br = new BufferedReader(new InputStreamReader(dis))) {
// words are separated by whitespace
Pattern pattern = Pattern.compile("\\s+");
String line = null;
while ((line = br.readLine()) != null) {
// do this if case sensitivity is not required i.e. Java = java
line = line.toLowerCase();
String[] words = pattern.split(line);
for (String word : words) {
if (wordMap.containsKey(word)) {
wordMap.put(word, (wordMap.get(word) + 1));
} else {
wordMap.put(word, 1);
}
}
}
} catch (IOException ioex) {
ioex.printStackTrace();
}
return wordMap;
}
public static List<Entry<String, Integer>> sortByValueInDecreasingOrder(Map<String, Integer> wordMap) {
Set<Entry<String, Integer>> entries = wordMap.entrySet();
List<Entry<String, Integer>> list = new ArrayList<>(entries);
Collections.sort(list, new Comparator<Map.Entry<String, Integer>>() {
@Override
public int compare(Map.Entry<String, Integer> o1, Map.Entry<String, Integer> o2) {
return (o2.getValue()).compareTo(o1.getValue());
}
});
return list;
}
}
Output:
List of repeated word from file and their count
its => 2
of => 2
programming => 2
java => 2
language => 2
6) How do you check if two given String are anagrams? (solution)
package miscellaneous;
/*
* We increment the count of each character in the first array and
* decrement the count of each character in the second array.
*
* If the resulting counts array is full of zeros, then strings are anagrams otherwise not.
*/
public class Find2StringAreAnagramsOfEachOther {
public static void main(String[] args) {
String str1 = "abc";
String str2 = "abc";
System.out.println("Anagram?? :"+isAnagram(str1, str2));
}
private static boolean isAnagram(String a, String b){
//if both string is null, then considering it as anagram.
if(a==b){
return true;
}
//if any one string is null, then they are not anagram.
if(a==null || b==null)
return false;
//If length of both strings are not same then obviously they are not anagrams.
if(a.length()!=b.length())
return false;
char[] aArr = a.toLowerCase().toCharArray();
char[] bArr = b.toLowerCase().toCharArray();
// An array to hold the number of occurrences of each character
int[] counts = new int[26];
for (int i = 0; i < aArr.length; i++){
counts[aArr[i]-97]++; // Increment the count of the character at respective position
counts[bArr[i]-97]--; // Decrement the count of the character at respective position
}
// If the strings are anagrams, then counts array will be full of zeros not otherwise
for (int i = 0; i<26; i++){
if (counts[i] != 0)
return false;
}
return true;
}
}
7) How to print all permutation of a String in Java? (solution)
package com.journaldev.string.permutation;
import java.util.HashSet;
import java.util.Set;
/**
* Java Program to find all permutations of a String
* @author pankaj
*
*/
public class StringHelper {
public static Set<String> permutationFinder(String str) {
Set<String> perm = new HashSet<String>();
//Handling error scenarios
if (str == null) {
return null;
} else if (str.length() == 0) {
perm.add("");
return perm;
}
char initial = str.charAt(0); // first character
String rem = str.substring(1); // Full string without first character
Set<String> words = permutationFinder(rem);
for (String strNew : words) {
for (int i = 0;i<=strNew.length();i++){
perm.add(charInsert(strNew, initial, i));
}
}
return perm;
}
public static String charInsert(String str, char c, int j) {
String begin = str.substring(0, j);
String end = str.substring(j);
return begin + c + end;
}
public static void main(String[] args) {
String s = "AAC";
String s1 = "ABC";
String s2 = "ABCD";
System.out.println("\nPermutations for " + s + " are: \n" + permutationFinder(s));
System.out.println("\nPermutations for " + s1 + " are: \n" + permutationFinder(s1));
System.out.println("\nPermutations for " + s2 + " are: \n" + permutationFinder(s2));
}
}
8 How do you print duplicate elements from an array in Java? (solution)
public class DuplicatesInArray
{
public static void main(String[] args)
{
String[] strArray = {"abc", "def", "mno", "xyz", "pqr", "xyz", "def"};
for (int i = 0; i < strArray.length-1; i++)
{
for (int j = i+1; j < strArray.length; j++)
{
if( (strArray[i].equals(strArray[j])) && (i != j) )
{
System.out.println("Duplicate Element is : "+strArray[j]);
}
}
}
}
}
9) How to convert String to int in Java? (solution)
Integer x = Integer.valueOf(str);
// or
int y = Integer.parseInt(str);
There is a slight difference between these methods:
• valueOf returns a new or cached instance of java.lang.Integer
• parseInt returns primitive int.
The same is for all cases: Short.valueOf/parseShort, Long.valueOf/parseLong, etc.
10) How to swap two integers without using temp variable? (solution)
a = a + b;
b = a - b;
a = a - b;
For strings (by using a temporary variable for the length):
int len1 = s1.length();
s1 = s1 + s2;
s2 = s1.substring(0, len1);
s1 = s1.substring(len1);
Categories:
Java interview Questions
0 comments: